// multiple of 8. A multiple of 8 has 3 low zero bits, so we just need to figure out how many
      // extra zero bits we need to add to the end of bitsPerChar to get 3 in total.
      // The logic here would be wrong for bitsPerChar > 8, but since we require distinct ASCII
      // characters that can't happen.
      int zeroesInBitsPerChar = Integer.numberOfTrailingZeros(bitsPerChar);